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If $R$ is a ring and $I$ is an ideal of $R$, then $F = R/I$ is a homomorphic image of $R$, i.e. there is a homomorphism $f: R \rightarrow F$.

If you let $M = F^n$, and define $(\cdot): (R,M) \rightarrow M$ as $a,(x_1,\ldots,x_n) \rightarrow (f(a)x_1f,\ldots,f(a)x_n)$ then $M$ is a module over $R$ because the associativity and distributivity of multiplication in $F$ carries over.

The case that I'm actually interested in is $F$ being a field (i.e., $I$ being irreducible and $R$ being a commutative ring with one). In that case, it seems that $M$ is nearly a vector space - you can, for example, for every $x,y \in M$, $a \in R$ with $x = ay$ find a $b \in R$ with $bx = y$.

I stumbled over this construction via the particular special case $M = GF(2)^n$ and $R = \mathbb{Z}$. Unfortunately, I haven't been able to find any reference to those kinds of modules - mostly because I don't really have a good idea what to search for. Is there a name for these things? Or can anyone point me to some literature?

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up vote 2 down vote accepted

Clearly $(R/I)^n$ is annihilated by $I$. Conversely, if $M$ is some $R$-module which is annihilated by an ideal $I$, then $M$ is canonically a module over $R/I$. If $R/I$ is a field, $M$ is free over $R/I$, i.e. isomorphic to $(R/I)^{(n)}$ for some cardinal number $n$.

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